(a) Let S be the distance covered by the particle between t = 0 tot = 10 s. Then,S = area under (v-t) graph = area of the triangle, whose base is 10 (s) andheight is 12 (m s $^{-1}$ ) $= \frac{1}{2} \times 10 \times 12 = 60 \text{ m}$Therefore, average speed = $\frac{\text{distance covered}}{\text{time taken}}$ $= \frac{60 \text{ m}}{10 \text{ s}} = 6 \text{ ms}^{-1}$(b) If $S_1$ and $S_2$ are the distances covered by the particle in the time intervals t = 2 s to 5 s and t = 5 s to 6 s, then the distance covered in the time interval t = 2 s to 6 s is given by $S = S_1 + S_2$To find $S_1$ : At t = 0, the particle is at rest (u = 0). Let $v_1$ be velocity of the particle after 2 s and a1 be the acceleration during the time interval $t_1$ = 0 s to 5 s. From figure, we have $a_{1} = \frac{12-0}{5-0} = 2.4 \text{ m} \text{ s}^{-2}$ $\therefore\; v_1 = u + a_1 t$ $= 0 + 2.4 \times 2 = 4.8 \text{ m} \text{ s}^{-1}$The distance $S_1$ is covered during the time interval t = 2 s to t = 5 s i.e. in time$t_1$ = 5 – 2 = 3 s and with the initial velocity $v_1$.Now, $S_1 = v_1 t_1 + \frac{1}{2} a_1 t_1^2$$\therefore\; S_1 = 4.8 \times 3 + \frac{1}{2} \times 2.4 \times 3^2$$= 14.4 + 10.8 = 25.2 \text{ m}$To find $S_2$ : Let $v_2$ be velocity of the particle after 5 s and $a_2$ be theacceleration during the interval t = 5 s to 10 s.From figure, we have$a_2 = \frac{0-12}{10-5} = -2.4 \text{ m} \text{ s}^{-2}$Also, $v_2 = 12 \text{ m} \text{ s}^{-1}$The distance $S_2$ is covered during the time interval t = 5 s to t = 6 s i.e. in time$t_2$ = 6 – 5 = 1 s and with initial velocity $v_2$.Thus, $S_2 = v_2 t_2 + \frac{1}{2} a_2 t_2^2$$S_2 = 12 \times 1 + \frac{1}{2}(-2.4) \times (1)^2$$= 12 - 1.2 = 10.8 \text{ m}$Hence, the required distance,$S = 25.2 + 10.8 = 36 \text{ m}$Also , average speed = $\frac{\text{distance covered}}{\text{time taken}}$ $= \frac{36}{6-2} = \frac{36}{4} = 9 \text{ m} \text{ s}^{-1}$